prove that radius of curvature =2*focal length
Answers
Explanation:
Consider a Concave mirror as shown in the figure above.
A ray of light AB travelling parallel to the principal axis PC is incident on a concave mirror at B. After reflection, it goes through the focus F. P is the pole of the mirror. C is the centre of curvature.
The distance PF=focal length f.
The distance PC=radius of curvature R of the mirror.
BC is the normal to the mirror at the point of incidence B.
∠ABC=∠CBF (Law of reflection, ∠i=∠r)
∠ABC=∠BCF (alternate angles)
⇒∠BCF=∠CBF
∴ΔFBC is an isosceles triangle.
Hence, sides BF=FC
For a small aperture of the mirror, the point B is very close to the point P,
⇒BF=PF
∴PF=FC=1/2PC
⇒f=1/2R
Now consider a Convex mirror as shown in the figure below.
A ray of light AB traveling parallel to the principal axis PC is incident on a convex mirror at B. After reflection, it goes to Dand appear to be coming from the focus F.
The distance PF= focal length f.
The distance PC=radius of curvature R of the mirror.
Straight line NBCis the normal to the mirror at the point of incidence B.
∠ABN=∠NBD(Law of reflection, ∠i=∠r)
∠CBF=∠DBN (vertically opposite angles)
∠NBA=∠BCF(corresponding angles)
⇒∠BCF=∠CBF
∴ΔFBC is an isosceles triangle.
Hence, sides BF=FC
For a small aperture of the mirror, the point Bis very close to the point P,
⇒BF=PF
∴PF=FC=1/2PC
⇒f=1/2R
Thus, for a spherical mirror (both for a concave and for convex), the focal length is half of radius of curvature.
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